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# lecture_39 (dragged) - MA 36600 LECTURE NOTES WEDNESDAY...

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Unformatted text preview: MA 36600 LECTURE NOTES: WEDNESDAY, APRIL 29 Nonhomogeneous Systems Constant Coeﬃcients. Consider now the nonhomogeneous system d x = A x + g(t) dt where A is a constant n × n matrix. We will show that the general solution to this system is in the form x(t) = x(c) (t) + x(p) (t) in terms of a homogeneous solution x(c) (t) = Ψ(t) c, where Ψ(t) = exp (A t); and a particular solution ￿t x(p) (t) = Ψ(t) Ψ(τ )−1 g(τ ) dτ . We use integrating factors, just like before. Recall the Product Rule for Matrices: d x = A x + g(t) dt exp (−A t) · d x − A x = g(t) dt ￿ ￿ d x + −A exp (−A t) · x = exp (−A t) g(t) dt ￿ ￿ d exp (−A t) · x = exp (−A t) g(t) dt ￿t exp (−A t) · x(t) = c + exp (−A τ ) g(τ ) dτ for some constant vector c. Now use the fact that Ψ(t) = exp (A t) Then we have the expression =⇒ Ψ(t)−1 = exp (−A t) . Ψ(t)−1 · x(t) = c + ￿ t Ψ(τ )−1 g(τ ) dτ . The claim follows when we multiply both sides by the fundamental matrix Ψ(t). We make a general remark that may help in ﬁnding this general solution x(t). Say for the moment that A has n distinct eigenvalues r1 , r2 , . . . , rn . Denote ξ (k) as an eigenvector corresponding to rk , and consider the matrices ξ11 ξ12 · · · ξ1n r1 0 · · · 0 ξ1k ξ21 ξ22 · · · ξ2n 0 r2 · · · 0 ξ2k T= . and D=. in terms of ξ (k ) = . . . . . .. . .. . . . . . . . . .. . . . . . . ξn1 ξn2 ··· 0 ξnn 0 ··· rn ξnk Then T−1 A T = D i.e., A is a diagonalizable matrix. Deﬁne the vectors y(t) = T−1 x(t) and h(t) = T−1 g(t). Then we ﬁnd the nonhomogeneous system d y = D y + h(t) dt =⇒ y(t) = Φ(t) c + Φ(t) 1 ￿ t Φ(τ )−1 h(τ ) dτ ...
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## This note was uploaded on 11/30/2011 for the course MATH 366 taught by Professor Edraygoins during the Spring '09 term at Purdue.

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